Question

A thin metal plate is placed against the teeth of a cog wheel. Cog wheel is rotated at a speed of 120 revolutions per minute and has 160 teeth.
(1) Calculate the frequency of notes produced.
(2) Calculate the speed of sound if the wavelength is 1.05m.
(3) What will be the effect when the speed of the cog wheel is doubled?

Answer 1

Hint: From the given number of revolutions per minute of the cog wheel we could find the number of revolutions of the cog wheel that is its frequency and multiplying this with number of teeth on the cog wheel gives you the frequency of the note produced. Recall the relation connecting speed, frequency and wavelength and then substitute the given values to get the speed of sound. Also, when the speed of sound is doubled there is an obvious increase in the frequency, that is the number of times the teeth of the cog wheel hitting the plate increase and thereby you could now determine the required effect.

Formula used: Expression for frequency,
$f=\dfrac{v}{\lambda }$

Complete step by step answer:
In the question, we are given a cog wheel against whose teeth a thin metal plate is kept. This cog wheel which has 160 teeth is being rotated at a speed of 120 revolutions per minute. We are asked to find the frequency of the note produced due to the teeth of the cog wheel striking against the plate, the speed of sound with the given wavelength and the effect when the speed of the cog wheel is doubled. So let us solve each subpart one by one.
(1) Frequency of the note produced:
Frequency by definition is the number of revolutions per unit time (one second). In the question speed of the cog wheel is given as 120 revolutions per minute. Therefore,
In one minute $\to $ 120 Revolutions
$\Rightarrow 60s\to 120$Revolutions, then,
$\Rightarrow 1s\to \dfrac{120}{60}$ Revolutions
$\Rightarrow 1s\to 2$ Revolutions
Therefore, the frequency of the cog wheel is,
$f=2Hz=2{{s}^{-1}}$
Frequency of the note produced is given by the number of times the teeth hits the plate per second. Clearly, it is given by the product of the number of teeth on the cog wheel and the frequency of the cog wheel.
${{f}_{n}}=n\times f$
${{f}_{n}}=160\times 2=320Hz$
Therefore the frequency of the note is 320Hz.
(2) Speed of sound for 1.05m wavelength:
We have the expression for frequency as,
$f=\dfrac{v}{\lambda }$ ………………………… (1)
$v=f\lambda $
Substituting the given values,
$\Rightarrow v=320\times 1.05$
$\Rightarrow v=336m{{s}^{-1}}$
Therefore, the speed of sound for 1.05m wavelength is $336m{{s}^{-1}}$ .
(3) Effect caused when the speed of the cog wheel is doubled:
When the speed of the cog wheel is doubled, by equation (1) there is subsequent increase in frequency, which means more number of revolutions per second, that is, the number of times the teeth of the cog wheel hits the plate increases thereby producing a high pitched sound.

Note: Remember that the cog wheel here has 160 teeth and the sound is produced when these teeth hit the plate so the frequency of the sound and the frequency at which the wheel is revolving is entirely different. So the increasing or decreasing the speed of the wheel leads to increase and decrease of the number of notes produced per second. As a result, a shrill sound is heard.